Whats the domain? someone pleaseee

Find the measure of the missing angles in the triangle below round your answer to the nearest degree

navik [9.2K]

Answer:

a. 40°

b. 21°

Step-by-step explanation:

a. Reference angle = θ

Opposite = 5

Adjacent = 6

Apply TOA:

$Tan (\theta) = \frac{Opp}{Adj}$

$Tan (\theta) = \frac{5}{6}$

$\theta = Tan^{-1}(\frac{5}{6})$

$\theta = 40$ (nearest degree)

b. Reference angle = x

Opposite = 5

Hypotenuse = 14

Apply SOH:

$Sin (x) = \frac{Opp}{Hyp}$

$Sin (x) = \frac{5}{14}$

$x = Sin^{-1}(\frac{5}{14})$

$x = 21$ (nearest degree)

3 0

2 years ago

the scale given on a road map is 1 cm equal 120 km what is the actual distance represented by 6 cm on the map

elena55 [62]

Answer:

720

Step-by-step explanation:

6 x 120=720

Every 1 cm=120 km

So multiply both sides by 6 and you get

6 cm= 720 km

8 0

2 years ago

Evaluate 23x + 4y when x=20 and y=23

never [62]

Answer:

552

Step-by-step explanation:

23x + 4y when x=20 and y=23

if x=20 and y=23 all you have to do is replace the numbers and solve

23(20) + 4(23)

Multiply 23*20 which is equal to 460

Then you multiply 4*23 which is equal to 92

The problem is now 460 + 92

add it, and you get 552.

Hope this helps!!

-Ketifa

3 0

3 years ago

Exactly 32% of the students in a school play a sport. Fifty students are randomly selected to determine the probability that, at

faltersainse [42]

A binomial probability density function should be used to represent the probability

<h3>How to determine the type of probability density?</h3>

The given parameters are:

Proportion that plays sport, p = 32%
Number of students selected, p = 50
The probability, P = (x ≤ 15)

The proportion that plays sport indicates that

68% of the students do not play sport

So, we have two events, which are

Play sport
Do not play sport

When there are two possible events, then the binomial probability density function should be used

Read more about binomial probability density at:

brainly.com/question/15246027

#SPJ1

8 0

2 years ago

If 2^n + 1 is an odd prime for some integer n, prove that n is a power of 2. (H

vovikov84 [41]

Step-by-step explanation:

We will prove by contradiction. Assume that $2^n + 1$ is an odd prime but n is not a power of 2. Then, there exists an odd prime number p such that $p\mid n$ . Then, for some integer $k\geq 1$ ,

$n=p\times k.$

Therefore

$2^n + 1=2^{p\times k} + 1=(2^{k})^p + 1^p.$

Here we will use the formula for the sum of odd powers, which states that, for $a,b\in \mathbb{R}$ and an odd positive number $n$ ,

$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2-...+b^{n-1})$

Applying this formula in 1) we obtain that

$2^n + 1=2^{p\times k} + 1=(2^{k})^p + 1^p=(2^k+1)(2^{k(p-1)}-2^{k(p-2)}+...-2^{k}+1)$ .

Then, as $2^k+1>1$ we have that $2^n+1$ is not a prime number, which is a contradiction.

In conclusion, if $2^n+1$ is an odd prime, then n must be a power of 2.

4 0

3 years ago